Find the probability that a number selected at random from the numbers 1, 2, 3, ..., 35 is a
(i)prime number
(ii)multiple of 7
(iii)a multiple of 3 or 5
Answers
SOLUTION :
Given : Numbers selected from 1 to 35
Total number of outcomes = 35
(i) Let E1 = Event of getting a prime number .
Numbers which is a prime number are = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31
Number of outcome favourable to E1 = 11
Probability (E1) = Number of favourable outcomes / Total number of outcomes
P(E1) = 11/35
Hence, the required probability of getting a prime number , P(E1) = 11/35 .
(ii) Let E2 = Event of getting a number which is a multiple of 7.
Numbers which is a multiple of 7 are = 7, 14, 21, 28, 35
Number of outcome favourable to E2 = 5
Probability (E2) = Number of favourable outcomes / Total number of outcomes
P(E2) = 5/35 = 1/7
Hence, the required probability of getting a multiple of 7 , P(E2) = 1/7 .
(iii) Let E3 = Event of getting a number which is a multiple of 3 & 5
Numbers which is a multiple of 3 & 5 are = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 5, 10, 20, 25, 35
Number of outcome favourable to E3 = 16
Probability (E3) = Number of favourable outcomes / Total number of outcomes
P(E3) = 16/35
Hence, the required probability of getting a multiple of 3 & 5 , P(E3) = 16/35 .
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Here's the answer:
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Given,
A number selected from 1 - 35
Let S be Sample Space
n(S) - No. of ways of selecting a number from 1 - 35
n(S) = 35C1 = 35
Let E1 be Event that selected number is a prime number
E1 = {prime No.s from 1 - 35}
E1 = {2, 3, 5 , 7, 11, 13, 17, 19, 23, 29, 31}
No. of favorable outcomes for occurrence of Event E1, n(E1) = no. of Elements in Set E1
n(E1) = 11
Let E2 be the Event that Number selected is a multiple of 7
E2 = { 7, 14, 21, 28, 35}
n(E2) = No. of multiplies of 7 from 1- 35
= 35/7
= 5
n(E2) = 5
No. of favorable outcomes for occurrence of Event E2, n(E2) = 5
Let E3 be the Event that a number selected is a multiple of 3 or 5
n(E3) = n{Multiples of 3} + n{Multiples of 5} - n(Multiplies of 3 & 5}
n(Multiplies of 3) = 35/3 = 11
[which are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33]
n(Multiplies of 5) = 35/5 = 7
[which are 5, 10, 15, 20, 25, 30, 35]
n(Multiplies of 3 & 5) = n(common multiplies of 3 and 5)
= 15, 30
= 2
•°• n(E3) = 11 + 7 - 2 = 16
Probability P(E) =
(i) P(E1) =
•°• Required Probability P(E1) =
(ii) P(E2) =
•°• Required Probability P(E2) =
(iii) P(E3) =
•°• Required Probability P(E3) =
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